Re: [math-fun] curious dyadic rationals t/b, b=2^f, t=sum of few +-2^e, ContinuedFraction(t/b) has 1's and 2's only

Hum, interesting but what is t ? what type of number is it ? what is the rule for t ? Simon Plouffe Le 29/09/2013 07:43, Warren D Smith a écrit :

As an example, consider

b = 2^63 = 9223372036854775808(decimal) = 8000000000000000(hexadecimal) = 1000000000000000000000000000000000000000000000000000000000000000(binary)

t = 5836946587753906177(dec) = 5100FFFF00000001(hex) = 101000100000000111111111111111100000000000000000000000000000001(binary)

ContinuedFraction(t/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2]

I have found that there are many such examples, and they are easy to find by computer search because t is a sum of few signed powers of 2. In the present example, t is a sum of 6 of them, and since binomial(64,6) * 2^5 = 2399179776 = 2.4 * 10^9 it would be feasible to consider all possibilities.

Another example has t' = 101000100000000111111111111111011111111111111111111111111111111(binary) which yields a very "nearby" continued fraction to the first example: ContinuedFraction(t'/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2]